Problem 42 Solve each equation. See Example... [FREE SOLUTION]

Chapter 10: Problem 42

Solve each equation. See Example $6 .$ $$ 2 x^{-2}-5 x^{-1}-3=0 $$

Short Answer

Expert verified

The solutions are $ x = \frac{1}{3} $ and $ x = -2 $.

Step by step solution

Substitute Variables

Let's make a substitution to simplify the equation. Let $ u = x^{-1} $. This means $ x^{-2} = u^2 $. Substitute these into the equation: $ 2u^2 - 5u - 3 = 0 $.

Solve the Quadratic Equation

Now we have a quadratic equation in the form $ 2u^2 - 5u - 3 = 0 $. Use the quadratic formula $ u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $, where $ a = 2 $, $ b = -5 $, and $ c = -3 $.

Apply Quadratic Formula

Calculate the discriminant: $ b^2 - 4ac = (-5)^2 - 4 = 25 + 24 = 49 $. The quadratic formula becomes $ u = \frac{5 \pm 7}{4} $.

Find Roots for u

The solutions for $ u $ are $ u = \frac{12}{4} = 3 $ and $ u = \frac{-2}{4} = -0.5 $. Thus, $ u = 3 $ and $ u = -0.5 $.

Substitute Back to x

Since $ u = x^{-1} $, we have $ x = \frac{1}{u} $. Thus, for $ u = 3 $, $ x = \frac{1}{3} $. For $ u = -0.5 $, $ x = \frac{1}{-0.5} = -2 $.

Verify the Solutions

Check both potential solutions by substituting back into the original equation to ensure no calculation error. Substitute $ x = \frac{1}{3} $ and $ x = -2 $ into the original equation to verify both are correct solutions.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Substitution Method

The substitution method is a powerful technique used to simplify complex equations. In the context of solving equations, it often involves introducing a new variable to replace an expression, making the equation easier to handle. Here's how it works:

Identify a part of the equation you can substitute with a simpler expression. In this problem, the term involving negative exponents, like $ x^{-1} $, is replaced with $ u $.
Write down what each substituted variable represents. For our exercise, we set $ u = x^{-1} $, which implies $ x^{-2} = u^2 $ since squaring both sides gives this relation.
Replace the identified parts in the entire equation with the substituted variables. Substituting $ u $ for $ x^{-1} $ transforms the equation from an expression with inverse powers of $ x $ to a quadratic equation: $ 2u^2 - 5u - 3 = 0 $.

The substitution step is crucial as it often transforms the equation into a much more recognizable form, like a quadratic equation, which can then be solved with standard methods.

Discriminant

The discriminant is a key concept in the realm of quadratic equations. It is a component of the quadratic formula and helps determine the nature of the roots of the quadratic equation. The formula for the discriminant is given by $ b^2 - 4ac $.

A positive discriminant ($ >0 $) indicates that the quadratic equation has two distinct real roots.
A zero discriminant ($ =0 $) suggests that the equation has exactly one real root, which is a repeated root, also known as a double root.
A negative discriminant ($ <0 $) means there are no real roots; instead, the roots are complex or imaginary.

In our exercise, after substituting and rearranging the equation, we calculated the discriminant as $ 49 $. This positive value assured us that there are indeed two distinct real roots for the quadratic equation.

Quadratic Formula

The quadratic formula is a universal method for finding the roots of any quadratic equation in the standard form $ ax^2 + bx + c = 0 $. It is expressed as $ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $. This formula derives from the process of completing the square and provides a straightforward solution for quadratic equations.

The quadratic formula directly uses the coefficients $ a $, $ b $, and $ c $ from the quadratic equation.
The "$ \pm $" symbol in the formula means it will produce two potential solutions, corresponding to the two roots of the quadratic equation when the discriminant is positive.
The discriminant $ b^2 - 4ac $ is under the square root. It's crucial to compute this first because it affects the type and number of solutions.

In our example, after determining the discriminant as 49, we used the quadratic formula to find $ u = \frac{5 \pm 7}{4} $, which resolved to the roots $ u = 3 $ and $ u = -0.5 $. This method reliably finds solutions, allowing us to substitute back and determine the original variable $ x $.

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91影视

Solve each equation. See Example \(6 .\) $$ 2 x^{-2}-5 x^{-1}-3=0 $$

Short Answer

Step by step solution

Substitute Variables

Solve the Quadratic Equation

Apply Quadratic Formula

Find Roots for u

Substitute Back to x

Verify the Solutions

Key Concepts

Substitution Method

Discriminant

Quadratic Formula

One App. One Place for Learning.

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